# Tag Info

29

Dyalog APL, 10 bytes (5?69),?26 Dyadic ? is ⍺ distinct random numbers in [1,⍵], and monadic ? is a single random number. Try it here.

27

MediaWiki templates with ParserFunctions, 48 bytes {{#ifexpr:1>{{#time:U}} mod {{{n}}}|true|false}}

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x86 machines with rdrand instruction, 10 bytes BITS 64 _try_again: rdrand eax jnc _try_again cmp eax, edi ja _try_again ret machine code 0FC7F0 73FB 39F8 77F7 C3 The input is in the register rdi and the output in rax. This respects the SYS V AMD64 ABI so the code effectively implement a C function unsigned int foo(unsigned int max); with 32-...

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Python A closed-form formula of p(n) is An exponential generating function of p(n) is where I_0(x) is the modified Bessel function of the first kind. Edit on 2015-06-11: - updated the Python code. Edit on 2015-06-13: - added a proof of the above formula. - fixed the time_limit. - added a PARI/GP code. Python def solve(): # straightforward ...

21

Jelly, 7 6 bytes ⁴!!X%‘ Thanks to @JonathanAllan for golfing off 1 byte! Cannot be run on TIO because (16!)! is a huge number. How it works ⁴!!X%‘ Main link. Argument: n ⁴ Set the return value to 16. !! Compute (16!)!. X Pseudo-randomly choose an integer between 1 and (16!)!. Since (16!)! is evenly divisible by any k ≤ 2**30, ...

15

Ruby 140 ->s{r=[1.0] s.lines.map{|l|n=[i=0.0]*(r.size+1) l.scan(/\S/).map{|e|a,b=e>?/?e>?]?[0.5]*2:[0,1]:[1,0] z=r[i] n[i]+=z*a n[i+=1]+=z*b} r=n} r} Function that takes as input the string (can be nicely formatted as a pyramid) and returns an array of floats. Test it online: http://ideone.com/kmsZMe Pretty straightforward implementation. Here ...

15

Pyth, 3 bytes !OQ Try it online Simple inversion of random choice from 0 to input Amusingly in Pyth it is not possible to make a function that does this without \$ because Pyth functions are automatically memoized.

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C++, 3477 3344 bytes Byte count does not include the unnecessary newlines. MD XF golfed off 133 bytes. There's no way C++ can compete for this, but I thought it would be fun to write a software renderer for the challenge. I tore out and golfed some chunks of GLM for the 3D math and used Xiaolin Wu's line algorithm for rasterization. The program outputs ...

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Python, 89 bytes from random import* lambda n,k:[x-i for i,x in enumerate(sorted(sample(range(1,n+k),k)))] Generating an increasing sequence rather than a non-decreasing one would be straightforward: this is just a random subset of k numbers between 1 and n, sorted. But, we can convert an increasing sequence to a non-decreasing one by shrinking each gap ...

13

Python 2, 269 Here's a nice little expression that evaluates to a function. It accepts three lists of integers. Passes all test cases. lambda A,B,C,w=lambda A,B:cmp(sum(cmp(a,b)for a in A for b in B),0),x=lambda A,B:cmp(sum(cmp(a+c,b+d)for a in A for b in B for c in A for d in B),0): (w(A,B)==w(B,C)==w(C,A)!=0)*((x(A,B)==x(B,C)==x(C,A))*["","strongly ","...

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Ruby, 140 158 bytes Don't keep upvoting this when there's a better ruby version. Here are more tricks for you. Unnamed function with one argument. Must not contain any spaces. May or may not contain a trailing newline. ->s{Z=(s.split' ')<<[] K=[] F=->i,j,f{k=Z[i][j] K[i]||=0 k==?^?2.times{|m|F[i+1,j+m,f/2]}:!k ?K[j]+=f :F[i+1,j+(k==?/?0:1),f]} F[...

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CJam, 16 bytes 69,mr5<26mr+:)S* 69, range(69) mr shuffle the generated array 5< first 5 elements 26mr random number 0..25 + concat to array :) increment each array element S* join with spaces Try it online.

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Pyth, 43 42 41 bytes umsdc+0sm@c[ZJhkJZKcJ2K)2x"\/"ekC,GH2.z]1 This expects the input to be without spaces. Try it online: Pyth Compiler/Executor Pyth, 40 bytes (questionable) umsdc+0sm,K@[ZJhkcJ2)x"\/"ek-JKC,GH2.z]1 Thanks to @isaacg, for saving one byte. Notice that this version didn't actually work in the version of Pyth, when the question was asked. ...

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CJam, 14 bytes E,_T9t\]ze~es= Test it here. Explanation E, e# Push [0 1 2 ... 12 13]. _ e# Make a copy. T9t\ e# Set the first element to 9. Swap with the original range. ]z e# Wrap them in an array and transpose to get [[9 0] [1 1] [2 2] ... [13 13]. e~ e# Run-length decode to get `[0 0 0 0 0 0 0 0 0 1 2 2 3 3 3 ... 13 13 ... 13 13]. es= e# Use ...

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CJam, 5 bytes Gotta be quick with these ones... rimr! Test it here. Explanation ri e# Read input and convert to integer N. mr e# Get a uniformly random value in [0 1 ... N-1]. ! e# Logical NOT, turns 0 into 1 and everything else into 0.

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TI-BASIC, 4 bytes using one byte tokens not(int(Ansrand Determines if the integer part of the input times a random number in [0,1) is zero. Ansrand<1 also works.

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Mathematica, 29 bytes Based on Dennis's Jelly answer. RandomInteger[2*^9!-1]~Mod~#& I wouldn't recommend actually running this. 2e9! is a pretty big number... It turns out to be shortest to generate a huge number that is divisible by all possible inputs and the map the result to the required range with a simple modulo. Rejection Sampling, 34 bytes ...

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Ruby, 204 191 172 characters c,*r=gets.split o=[0]*8 s=->x,y,p{y>14?o[x]+=p :(r.index("#{x},#{y+=1}")?(x<1?s[x+1,y,p]:(x>6?s[x-1,y,p]:(s[x-1,y,p*0.55]+s[x+1,y,p*0.45]))):s[x,y,p])} s[4,0,1] p o[c.to_i] It recursively simulates all possibly outcomes while keeping track of each individual outcome's probability, then it adds that probability to a ...

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MATL, 5 bytes Three different versions of this one, all length 5. iYr1= which takes an input (i), generates a random integer between 1 and that number (Yr), and sees if it it is equal to 1 (1=). Alternatively, li/r> make a 1 (l, a workaround because there is a bug with doing 1i at the moment), take an input (i), divide to get 1/N (/), make a random ...

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Python 3.3+, 64 bytes import math lambda s:sum(math.log2(len(s)/s.count(c))for c in s) Got math.log2 from mbomb007's solution.

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Python 3, score = big(?) from math import exp, log, log1p def f(b, n): e = n * (n - 1) / 2 m = 0 c = 1 s = 0 t = 1 << b for k in range(b): s += c m += exp(e * (log1p(-s / t) if 2 * s < t else log((t - s) / t))) c = c * (b - k) // (k + 1) return m Try it online! The Hamming distance $D_{x, y}$ ...

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CJam, 50 48 45 44 42 40 bytes 1]q{iD%"(+0 0+( (\Y/::+ (d2/_a+"S/=~+}/p This expects the input to be without space and have a trailing newline. For example: ^ \^ ^^\ \^/^ [0 0.1875 0.5625 0.125 0.125] Algorithm The basic idea is that you keep on parsing each character (there are only 4 different characters) and perform different operations on the ...

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Julia, 17 16 15 bytes n->2>rand(1:n) This is a function that generates a random integer between 1 and n and tests whether it's less than 2. There will be a 1/n chance of this happening, and thus a 1/n chance of returning true. Saved 1 byte thanks to Thomas Kwa!

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JavaScript ES6, 15 bytes -5 bytes thanks to Downgoat. x=>1>new Date%x Based off (uses) of this answer's technique.

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Python Note: Congratulations to Min_25 for finding a closed-form solution! Thanks for the interesting problem! It can be solved using DP, although I am not currently feeling very motivated to optimise for speed to get a higher score. It could be nice for golf. The code reached N=39 within 10 seconds on this old laptop running Python 2.7.5. from time ...

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Microscript II, 3 bytes NR! Reads an integer n, generates a random integer between 0 and n-1 (inclusive), then applies a boolean negation to that value.

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MATL, 10 bytes 69 5Zr26Yr Uses current version (9.2.0) of the language/compiler. Example With compiler run on Matlab: >> matl > 69 5Zr26Yr > 66 59 64 56 29 12 With compiler run on Octave: >> matl > 69 5Zr26Yr > 2 69 41 44 23 22 The first five numbers are separated by space, not by newline. This is due to Octave's underlying ...

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Mathematica, 29 bytes BetaRegularized[#3,#,1+#2-#]& Takes input in the order n,m,p. Mathematica is so good, it even golfs your code for you: BetaRegularized is the regularised incomplete beta function.

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Brachylog, 9 bytes ≥.∧13ḟṙ|↰ Try it online! This uses 13! like in Martin Ender's answer (13ḟ is one byte less than 2^₃₀). ṙ is implemented using random_between/3, which, when digging its source, uses random_float/0 which is linked to random/1 which uses the Mersenne Twister algorithm which is uniform for our purposes. Explanation ≥. Input ≥ Output ...

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Mathematica, 47 bytes Plot3D[E^(-(x^2+y^2)/2/#^2),{x,-6,6},{y,-6,6}]& takes as input σ Input [2] output -2 bytes thanks to LLlAMnYP

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